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Solved Examples – Two and Three Dimensional Force Systems

SOLVED EXAMPLES

Example 3.1. The force F has a magnitude of 300 N. Express F as a vector in terms of

the  unit vectors i and j. Identify the x and y scalar components of F.

Solution .   f  = – 300 sin 30 0  +300 cos 300 j

= - 150 I +259.81 j N

FX =150 N

FY =259.81 N 

 

Example 3.2. The slope of the 150 N force is specified as shown in Fig. P-3.2. Express F as a vector in terms of the unit vectors i and J.

Solution. Unit vector along F = 

F =3i -4j/5 x 150 N

= – 90 I -120 j N

 

Example 3.3. A cable stretched between the fixed supports A and B is under a tension T of90 N. Express the tension as a vector using unit vectors i and J, first, as a force T; acting on A and second, as a force T,; acting on B.

Solution .  

Example 3.3. A cable stretched between the fixed supports A and B is under a tension T of  90 N. Express the tension as a vector using unit vectors i and J, first, as a force T; A  acting on A and second, as a force T,; acting on B.

Solution : 

Unit vector along T A =

 

Example 3.4. To satisfy design limitations it is necessary to determine the effect of the 20 k N tension in the cable on the shear, tension, and bending of the fixed !-beam. For this purpose, replace this force by its equivalent of two forces at A, F; parallel and F,; perpendicular to the beam. Determine F1 and F,.

Solution :              f t = 20 cos  (200 + 300)

= 12.855 kN

Fn =20 sin (200 +300)

= 15.32 KN .

 

Example 3.5. If the equal tensions T in the pulley cable exert a force R on the pulley, determine the vector R and its magnitude in terms of T.

 

 

Example 3.6. The circular cam has an offset θ = 25 mm and a radius r =50 mm. For the position where θ = 30°, the smooth undersurface of the plunger exerts a downward force of 160 N on the cam normal to the contacting surfaces. In designing the cam bearing it is necessary to calculate the rectangular component F1 of this force along the line joining the contact point and the center of the shaft which turns the cam. Find F1

Solution.

In triangle OCB, we have LOBC + α = 90°-30° = 60°

β +a = 60°.

Again  50  /sin β  25 /sin α

Or                                2 sin α = sin β

or                                  2sin(60°-P)=sin β

or                                 2 [ √ 3/2  cos β – 1/2  sin β] = sin β

√3 cos β – sin β = sin β

Tan β = √3 /-2

β= 40.89°

α= 19.11°.

Force along                 CO= 160 N

. . Rectangular component along CB

= 160 x cos α

= 160 x cos 19.11 0

F1 = 151.182 N .

Example 3.7. The 12-kN force F is applied at point A. Compute the moment of F about point 0, expressing it both as a scalar and as a vector quantity. Specify the x-coordinate of the point B on the x axis about which the moment of F is zero. 

Solution.              Unit vector along F

CCW  =   counter clockwise

=41.46 k nm (CCW)Vector

| M | =41.46 KN (CCW) . (SCALAR)

MB =0 it means F ‘s line of action passes through B.

Tan θ =5/12

5/12 =15 /27+xB

XB +27 =36

X B =9 M

Example 3.8. The rectangular plate is made up of 0.3 m squares. as shown. A 15 N force is applied at point A in the direction shown. Calculate the moment M; of the force in scalar about point B.

Solution

Unit vector along F=

M B =R BA x f

= – 12.48 K

à

| M | = 12. 48 CW .

 

Example 3.9. The wooden crate is resting on the right-angled frame that is pivoted at 0 and supported at A. The force F must supply a moment about 0 of 12 kNm in order to reduce the support force at A to zero. Determine F.

Solution.                  

 

Taking moment about 0 we get,

M0 =F sin 300 X 0.8  +F cos 300 X 0.5 (CW)

GIVEN                       M0 =12 KN m

OR                              F sin 300 x 0.8 +f cos 300 x 0.5  =12

Or                                f x 0.4 +f x √3/2  =12

F =14.4 KN

D m A /d θ = 1229 .1 cos θ -509 .1 sin θ =0 for MA (max)

Tan θ =1229.1 /509 .1 è θ 67.5 0.

MA (max) =1229.1 sin 67.5 +509 .1 cos 67.5 =1330 .36 N .m

 

Example 3.11. Calculate the combined moment of the two 40 N forces about points 0 and A respectively. Comment on the results.

Solution.

 MO = – 40 x 0.2 -40 x (0.2)

= – 16 N m

Negative sign shows clockwise moment

Comment : 40 N and 40 N forces constitute a couple. The moment of the couple is M = 40 x (0.2 + 0.1 + 0.1) or M = 16 Nm clockwise. So the couple moment is fixed it is 16 Nm clockwise, it does not depend on point 0 or A.

M= Mo=MA

 

Example 3.12. Replace the force-couple system at point 0 by a single force. Specify the coordinate of the point on the y-axis through which the line of action of this resultant force passes.

Solution.

Couple and· force is shifted to point A by a single force F . f ‘ and- F constitute the given couple 8 kN.

. . anticlockwise moment (f x YA ) = – 8

YA -8/20 = – 0.4

Y A = – 0.4 m

 

Example 3.13. The indicated force-couple system is applied to a small shaft at the centre of the square plate. Replace this system by a single force and specify the coordinate on y axis through which the line of action of this resultant force passes. What would be the maximum magnitude of M for which the line of action of the equivalent resultant force would be located on the plate rather than on an extension of the plate ? (Ref Fig. P-3.13(a))

Solution.

300 x Ya =7000

YA =7000 /300 =70/3 m =23.33 m

Maximum moment would be at y B = 30 m

MB = 300 x 30 = 9000 N-m. Ans.

 

Example 3.14. Determine and locate the resultant R of the two forces and one couple acting as shown in Fig. P-3.14(a). 

Solution.

Ʃ F y =7 -10 = -3 KN

R =3 KN downward .

Me (for all forces and couple)

= – 7x 2 + 38 -10 x 2

= – 14 + 38 -20 =4 knw (ccw )

R (4- ) =4

3 (4 –x  ) =4

12 -3x =4

X 8/3 =2.67 m

 

Example 3.15. Express the 90 N force F as a vector in terms of the unit vectors i, j and k.  Determine the projection of F onto the y-axis. 

Solution .                            

Unit vector along     AB   =

 

Example 3.16. A vector A is 3i -2 j + 2k. Find the projection of A on the line joining B (1, 2, -3) and C (-1, -2, -2) directed to B to C. 

Solution. Unit vector along BC =

Thus the projection of A on BC is

= unit

vector along BC

= -6 +8+2 / √21 = 4/ √21 = 4 4.5826 =  0.873

 

Example 3.17. Derive the expression for the projection of the force F onto the line directed from D to C. 

Solution.

Unit vector along D to C is

Projection of F on DC (D to C),

= Dot product between F and unit vector along DC.

 

Example 3.18. Replace the two forces that act on the 3 m cube by an equivalent single force F at A and a couple. 

Solution.                          F1 =300  (-J) N

R (Resultant force) =F 1  +F2                        

Equivalent force at A= R = .

Moment of 300 N about A

Momentof400Nabout A=

 

Example 3.19. Determine the wrench resultant of the three forces acting on the bracket. Calculate the coordinates of the point P (x, y) through which the resultant force of the wrench acts. Find also the magnitude of the couple M of the wrench.

Solution. Let x and y be in mm

F1  acting along y at a

F2 acting along x at b

F3 acting along x at c.

Resultant force

The direction cosines of F are

Cos θ x 2/6 =1/3 ,cos θ y =4/6 =2/3 ,cos θ z =4/6 =2/3

Moment of F2 about   p= y j x 2 I =2y k n mm

Moment of F1 about  p= (-y j –x I +60 k ) x  j n-mm

Moment of F2 ab0ut  p = [(80 –y ) j + (100 –x) I ] x 4 kn=mm

Sum of all moments

For wrench resultant principle, “the direction cosines of the couple M of the wrench

must be the same as those of the resultant force F, assuming that the wrench is positive

thus ,                           cos θ x =1/3 =80 -4y /m

cos θ y =2/3 = – 400 +4x  /m

cos θ z = 2/3 = -4x +2y /m

There are three unknowns M, x and y and three equations, so solving we get

x = 60 mm, y = 40 mm, M =- 240 N-mm.

M is negative, which implies that the wrench is negative.